Answer:
Final temperature of the ice tea, x = 42.73°C
Explanation:
230 cm^3 = 230 g of liquid ( taking the density of tea as the same the density of water)
In cooling from 96°C to 0°
C (before freezing) the available energy is
96 × 230 = 22080 cal
The latent heat (the heat required to melt the crushed ice per gram) of fusion for water = 79.7 cal/g
Melting 100 g of crushed ice at 0°C will absorb
100 × 79.7 = 7970 cal
The retain heat of the iced Tea = (22080 - 7970) cal of heat raising the mixture temperature as follows
(22080 - 7970) = 14110 cal
The mass of the final solution 230 + 100 = 330 g solution
Final temperature = 14110 / 330 = 42.76°C
Or where specific heat of water = 4.186J/Kg°C
and the latent heat of the ice = 334J/Kg
We have
230 × (specific heat capacity of water) × (96 - x) =100× (specific heat capacity of water) ×(0-x) + (latent heat) ×100 =
230 × 4.186 × (96 - x) =100× 4.186 ×(x-0) + 100×334
1381.38×x = 59026.88
Final temperature of the ice tea, x = 42.73°C
